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I was thinking about the complex unit circle $S_1\subset\mathbb{C}$ as a group under multiplication and how, if possible, an element $z$ of infinite order could generate the circle itself. A friend quickly pointed out that $z$ cannot be algebraic, since there are points on $S_1$ with transcendental argument which cannot, by definition, be expressed as some polynomial of algebraic numbers. After mulling this over for a couple of minutes, I formulated the following question:

Let $\tau$ be some fixed transcendental number. Can any trancendental number $\sigma$ be represented as $\tau^n + m$, for some $n\in\mathbb{N}, m\in\mathbb{Z}$?

Justin
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  • @2000 I'm not quite sure what you're asking. – Justin Mar 06 '16 at 10:18
  • @2000 I can't understand your question. The asker seems to be asking whether any transcendental number is some natural power of some fixed transcendental added to some integer, like fix $;\pi;$ , then for example is it possible to write $;e=\pi^n+m;$ for some $;n\in\Bbb N;,;;m\in\Bbb Z;$ ? – DonAntonio Mar 06 '16 at 10:50
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    OK , now I understand your question. But the answer is no. Because we have uncountable trancendental numbers. –  Mar 06 '16 at 10:54
  • See Hamel Basis . –  Mar 06 '16 at 11:48

1 Answers1

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The group $S^1$ is not cyclic because it is uncountable. The set $\{\tau^n +m \mid n,m \in \Bbb Z\}$ is countable, hence it cannot be equal to the (uncountable) set of transcendental numbers.

However, you can say that $\tau^n + m$ is transcendental if $n≠0, m \in \Bbb Z$.

By Lindemann–Weierstrass theorem, $\tau=e$ and $\sigma = e^{\sqrt 2}$ are algebraically independent (and $\sigma$ is transcendental by the same theorem). In particular, you cannot write $\sigma = \tau^n + m$ for some integers $n$ and $m$.

[I don't known whether it is possible to write $e=π^n+m$ for some integers $n$ and $m$. I believe that it is still unknown ; at least the algebraic independence of $e$ and $π$ is unknown.]

Watson
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